Proving Module Homomorphisms: A x B to M & M to A x B

In summary, the conversation discusses two isomorphisms involving HomR, a commutative ring R, and R-modules A, B, and M. The first isomorphism is between HomR(A x B, M) and HomR(A, M) x HomR(B, M), and the second isomorphism is between HomR(M, A x B) and HomR(M, A) x HomR(M, B). The conversation also mentions constructing homomorphisms between M and A or B given a homomorphism from M to A x B, and vice versa.
  • #1
JdotAckdot
4
0
If you can help, that would be great.

Let R be a commutative ring, and A,B,M be R-modules. Prove:

a) HomR(A x B, M) is isomorphic to HomR(A, M) x HomR(B, M)
b) HomR(M, A x B) is isomorphic to HomR(M, A) x HomR(M, B)
 
Physics news on Phys.org
  • #2
step one: try to write down a map from one to the other.

e.g. given a pair of homomorphisms f:M-->A and g:M-->B. how would you construct, in the simplest most natural way, a homomorphism M-->AxB?

conversely, given a homomorphism M-->AxB, how would you construct homomorphisms M-->A and M-->B?
 
  • #3


a) To prove that HomR(A x B, M) is isomorphic to HomR(A, M) x HomR(B, M), we need to show that there exists a bijective homomorphism between the two sets. Let f: HomR(A x B, M) -> HomR(A, M) x HomR(B, M) be defined as f(phi) = (phi1, phi2) where phi1(a) = phi(a, 0) and phi2(b) = phi(0, b) for all a in A and b in B. It is clear that f is a homomorphism since for any (a,b) in A x B, we have f(phi)(a,b) = (phi1(a), phi2(b)) = (phi(a,0), phi(0,b)) = phi(a,b). Thus, f is a homomorphism.

To show that f is bijective, we will show that f is injective and surjective. To prove injectivity, let phi1, phi2 be two elements in HomR(A, M) and HomR(B, M) respectively such that f(phi1) = f(phi2). This means that phi1(a) = phi2(a) and phi1(b) = phi2(b) for all a in A and b in B. Then, for any (a,b) in A x B, we have f(phi1)(a,b) = (phi1(a), phi1(b)) = (phi2(a), phi2(b)) = f(phi2)(a,b). Thus, phi1 = phi2 and f is injective.

To prove surjectivity, let (phi1, phi2) be an element in HomR(A, M) x HomR(B, M). Then, for any (a,b) in A x B, we have f(phi1, phi2)(a,b) = (phi1(a), phi2(b)) = (phi(a,0), phi(0,b)) = phi(a,b). Thus, f is surjective and hence bijective.

Therefore, HomR(A x B, M) is isomorphic to HomR(A, M) x HomR(B, M).

b) To prove that HomR(M, A x B) is isomorphic to HomR(M, A) x HomR(M, B), we will use a similar
 

Related to Proving Module Homomorphisms: A x B to M & M to A x B

What is a module homomorphism?

A module homomorphism is a function that preserves the algebraic structure between two modules. In this case, the function maps elements from the direct product of two modules, A x B, to the direct product of M and M, and vice versa.

What is the purpose of proving module homomorphisms?

The purpose of proving module homomorphisms is to establish a mathematical relationship between two modules that allows for easier computation and understanding of their properties. It also allows for the generalization of concepts and the ability to apply them to different modules.

Why is it important to prove the homomorphism in both directions?

Proving the homomorphism in both directions ensures that the function is bijective, meaning it is both injective and surjective. This guarantees that the mapping between the two modules is well-defined and allows for the inverse function to exist.

What techniques are commonly used to prove module homomorphisms?

The most commonly used technique to prove module homomorphisms is by using the properties of the modules, such as linearity and associativity. Other techniques include using the definition of a homomorphism and using the properties of the underlying ring or field of the modules.

What are some applications of module homomorphisms in science?

Module homomorphisms have various applications in science, particularly in abstract algebra and its applications in fields such as physics, chemistry, and engineering. They are also used in signal processing, coding theory, and cryptography. In addition, module homomorphisms are essential in the study of group representations and commutative algebra.

Similar threads

  • Linear and Abstract Algebra
Replies
13
Views
2K
  • Linear and Abstract Algebra
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
1K
  • Math POTW for University Students
Replies
7
Views
1K
  • Linear and Abstract Algebra
Replies
7
Views
2K
  • Linear and Abstract Algebra
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
4
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Math POTW for Graduate Students
Replies
2
Views
736
  • Linear and Abstract Algebra
Replies
1
Views
1K
Back
Top